I see, so for example, you want to prove that, by using only the operations and functions in B2, the number of 1's required to make any fraction is the same as the number of 1's required to make that fraction, using only the things in B3 instead. I think you need to prove this, in some sense, by induction. Assume that you can create the numbers x and y using either B2 or B3 with the same number of 1's. Now, using B2, you can create x + y using no more 1's, and the same is true for B3. You can create x-1 and y-1 using no more 1's with B2, so can the same be said for B3? We'll get back to that. Using one more 1, you can create x+1, y+1, 1-1, and 1 using B2, and the same is obviously true for B3. Using two more 1's, you can create 1+1 using B2, and the same is true for B3.
Using no more 1's with B3, you can create x+y and (x-1 + y-1)-1 and using B2 you can obviously create x+y using no more 1's and you can create the second expression using no more one's as well (look at it for a second, it will become obvious). Using one more 1 with B3, you can create 1, x+1, y+1, (x-1 + 1-1)-1, (1-1 + y-1)-1, and if you think about it a little, you'll see that you can do these things with also just one more 1 using B2. Using two more 1's, you can create 1+1, and (1-1 + 1-1)-1 = 1/2, and again, you'll see that this can easily be done with also using just two more 1's with B2.
So getting back to the one we missed? Can we invert numbers using B3 without using any more 1's? Actually, I don't think it can be done. Using x-1 with B2 only requires you to have x occur once. The formulas that B3 has requires at least two things to appear, so it will require the 1's from two different things, and assuming one of those things is x, given that the other one will contain at least one 1, you have to use at least one more 1 to create x-1. Now, this assumes that the best way to construct x-1 is to create x, then invert it, and if what we're required to prove is indeed true, then this can't be the case for B3. Now clearly, the "quickest" (that is, the way using least 1's) way to create x-1 with B2 is to create x and then invert it. Suppose it takes n 1's to create x, then it creates n 1's to create x-1 if you create x and then invert it. If there is a quicker way to get directly to x-1 that takes m < n 1's, then it only takes m 1's to create (x-1)-1 = x. We assumed n to be the least number of 1's required to make x, then assumed that there could be a quicker way to make x-1, and found that this implies that there is some m < n such that you can create x with only m 1's, that is, there is a way quicker than the quickest, contradiction. So there is no quicker "direct" way to make x-1, you have to just make x and then invert it. What we want to show, however, that this is not the case for B3. That there is some way to make x-1 without making x first. We also want to show that the number of 1's required to make x (and hence the number of 1's required to make x-1) using B2 is the same as the number of 1's required to make x-1 using B3.
Actually, since the number of 1's required to make x is the same as the 1's required to make x-1 using B2, the above is equivalent to asking that we prove that it takes the same amount of 1's to create x-1 with B2 as it does for B3, and since x is arbitrary, we may as well say that it takes the same number of 1's to create x using B2 or B3. This is actually what we wanted to prove in the first place, so we're back at the beginning. Note that now, however, we're not assuming anything specific about x (we've left the scope of the assumption that B2 and B3 realize x with the same number of 1's).
Maybe we can go back to assuming that x takes the same 1's with B2 or B3, and still prove that x-1 takes the same 1's using B3. Note that for B2 to make x, it had to have added two previous numbers w and z. Let's assume that B3 can make w and z with as many 1's as B2 (this is consistent with our assumption because it then takes B3 no more 1's to add w and z together, just as B2 would do, however, this doesn't follow from our assumption, it's an additional one). Can we use B3 to take w and z and jump straight to x-1 without using any more 1's? Actually, I don't see this working, but I've thrown some ideas out, maybe someone can build on them (note that using B3's second operation, you just get zw/(z + w), but we need 1/(z + w)).
These ideas are very interesting 
